An object is falling freely under gravity. How much distance will it travel 2nd and 3rd second of its journey? (Ans. 15 m, 25 m)


Given:


Initial Velocity of object = vi = 0 m/s ( Let the object start falling from rest) and taking the value of acceleration due to gravity = g = 10 m/s²
 

To Find:

Distance covered in 2nd-second journey = d₂ = ?
Distance covered in 3rd-second journey = d₃ = ?

Solution:

According to the given data, we will use the second equation of motion (general form for a free falling objects)

S = vi t + 1/2 g t² ----------(1)

Using equation (1)

The distance travelled after 1st second ( t = 1 s ) ;

S1 = vi t + 1/2 a t²

S1 = 0 m/s ｘ 1s + 1/2 ｘ 10 m/s² (1 s)² = 0 + 5 m = 5m


The distance travelled after 2nd second ( t = 2 s ) ;

S2 = (vi t + 1/2 a t² )

S2 = 0 m/s ｘ 2s + 1/2 ｘ 10 m/s² (2 s)² = 0 + 25 m = 25 m


The distance travelled after 3rd second ( t = 3 s ) ;

S3 = vi t + 1/2 a t²

S3 = 0 m/s ｘ 3s + 1/2 ｘ 10 m/s² (3 s)² = 0 + 45 m = 45 m


The distance covered in 2nd-second journey is;

d2 = S2 - S1 = 20 m - 5 m = 15 m


Similarly, the distance covered in 3rd-second journey is;

d3 = S3 - S2 = 45 m - 20 m = 25 m



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